# Find the general solution of $4u_{xx}+5u_{xy}+u_{yy}+u_x+u_y=2$

How do I solve the following second order partial differential equation?

$4u_{xx}+5u_{xy}+u_{yy}+u_x+u_y=2$

I have classified the equation to be hyperbolic and changed variables to obtain the canonical form as $u_{\epsilon\nu}={1\over{3u_\nu}}-{8\over9}$ which I believe is correct but I am struggling to find the general solution? A step by step solution would be appreciated.

• If your canonical form is correct and you have $$u_{\eta \xi} - \frac{1}{3} u_{\xi} = -\frac{8}{9}$$ then integrating with respect to $\xi$ gives $$u_{\eta} - \frac{1}{3} u = -\frac{8}{9} \xi + f(\eta)$$ Now just use an integrating factor $$(e^{- \eta / 3} u)_{\eta} = e^{- \eta / 3} \left ( -\frac{8}{9} \xi + f(\eta) \right)$$ Hopefully you can take it from there. – mattos Nov 4 '17 at 15:29

• $4D_x^2+5D_xD_y+D_y^2+D_x+D_y=(D_x+D_y)(4D_x+D_y+1)$
• Solution of $u_x+u_y=0\;$ is $\;u_1=f_1(x-y)$
• Solution of $4u_x+u_y+u=0\;$ is $\;u_2=e^{-\frac{x}{4}}f_2(x-4y)$
• Particular solution is $\;u_p=2x$
• Answer: $u=u_1+u_2+u_p=f_1(x-y)+e^{-\frac{x}{4}}f_2(x-4y)+2x$